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Question: An arithmetic expression consists of 20 terms of which \({{4}^{th}}\) term is 16 and the last term i...

An arithmetic expression consists of 20 terms of which 4th{{4}^{th}} term is 16 and the last term is 208. Find the 15th{{15}^{th}} term.
(a) 146
(b) 147
(c) 148
(d) 149

Explanation

Solution

We can find the 15th{{15}^{th}} term of the arithmetic progression by first finding the common difference (d) of the arithmetic series. We have to form two simultaneous linear equations, one with the 4th term as the nth term and the other with the 20th{{20}^{th}} term as the nth{{n}^{th}} term using the formula an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d and solve both the equations simultaneously.

Complete step-by-step answer:
We first need to evaluate the formula for Arithmetic progression,
Suppose a1,a2,a3,..........,an{{a}_{1}},{{a}_{2}},{{a}_{3}},..........,{{a}_{n}} is an AP, then; the common difference “d” can be obtained as;
d=a2−a1=a3−a2=.......=an−an−1d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=.......={{a}_{n}}-{{a}_{n-1}}

The formula for finding the nth{{n}^{th}} term of an AP is:
an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d
Where,
a = First term
d = Common difference
n = number of terms
an=nth term{{a}_{n}}={{\text{n}}^{th}}\text{ term}
Let us now consider the given question,

It is given that,
n=20 a4=16; a20=208 \begin{aligned} & n=20 \\\ & {{a}_{4}}=16; \\\ & {{a}_{20}}=208 \\\ \end{aligned}
Applying the nth term formula,
an=a+(n−1)d a4=a+(4−1)d \begin{aligned} & {{a}_{n}}=a+\left( n-1 \right)d \\\ & {{a}_{4}}=a+\left( 4-1 \right)d \\\ \end{aligned}
a4=a+3d{{a}_{4}}=a+3d…………… (1)
a20=a+(20−1)d ⇒208=a+19d........(2) \begin{aligned} & {{a}_{20}}=a+\left( 20-1 \right)d \\\ & \Rightarrow 208=a+19d........(2) \\\ \end{aligned}

Now let us subtract equation 2 from equation 1.
−192=−16d ⇒19216=d \begin{aligned} & -192=-16d \\\ & \Rightarrow \dfrac{192}{16}=d \\\ \end{aligned}

Now, let us substitute the value of d in equation (1).
16=a+3(12) ⇒16−36=a ⇒a=−20 \begin{aligned} & 16=a+3\left( 12 \right) \\\ & \Rightarrow 16-36=a \\\ & \Rightarrow a=-20 \\\ \end{aligned}

Now to find the value of the 15th{{15}^{th}} term,
a15=a+(15−1)d ⇒a15=−20+14(12) ⇒a15=−20+168=148 \begin{aligned} & {{a}_{15}}=a+\left( 15-1 \right)d \\\ & \Rightarrow {{a}_{15}}=-20+14\left( 12 \right) \\\ & \Rightarrow {{a}_{15}}=-20+168=148 \\\ \end{aligned}
Therefore, the value of the 15th{{15}^{th}} term is 148.

Hence, the final answer is option (c).

Note: For solving any problem related to arithmetic progression, we need to have knowledge of 3 things namely first term (a), number of terms (n) and common difference (d). Once, we know these 3 terms, one just needs to apply the constraints given in the problem to get the required answer.